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A level Biology: Post your doubts here!

qwertypoiu

qwertypoiu

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Studydayandnight said:
Also this one! Correct answer is C again.
Click to expand...
In vascular bundle in transverse section of leaf, xylem is always on top, and phloem at bottom.
So that negates B and D.
Between C and D, the difference is in the relative widths of palisade and spongy mesophyll, it is quite clear that the diagram C has the bigger relative width for palisade layer as the question requires.
Revise leaf structure:
structure-of-leaves.jpg
 
Mahnoorfatima

Mahnoorfatima

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There's this question where it says: Explain how it is possible to reduce the no. of deaths during a cholera epidemic in countries such as those in W.Afrrica. In the ER it says that writing about washing hands and stuff is inappropriate. But why? I wrote that people who're infected should be advised not to handle cooking or utensils so that the bacteria doesn't go into the food and infects uninfected people or if they do, the should wash their hands first? Is this wrong? And also that proper sanitation facilities should be provided so that the bacteria doesn't remain in the faeces and infects other people. THIS IS DISGUSTING. :sick:
 
qwertypoiu

qwertypoiu

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ashcull14 said:
View attachment 52700
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The only thing that can denature protein is very high temperature or change in pH, based on what we learnt about enzymes. 40 degrees is not high enough.
So I'd go for C, since even for phospholipids dissolving the best option seems to be ethanol (probably due to their long fatty acid chains), other options are not plausible.
 
qwertypoiu

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ashcull14 said:
View attachment 52702
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Imagine there are 5 enzymes inside a solution.
Now if there was only one substrate molecule, after a while it will meet an enzyme and will form products.
Increasing the substrate concentration increases the number of molecules of the substrate per unit volume, thus increasing the number of likely collisions that happen in a given time initially, and thus increasing the rate of reaction. So, increasing substrate concentration increases the reaction rate. However, past a certain point the number of substrate molecules increases too much and the reaction rate no longer increases, this is because all five of the enzymes are saturated with a substrate to work on.

Now the inhibition part. A non-competitive inhibitor will just come in, and block let's say two of the enzymes. This means, when there is 1 substrate molecule, there is only 3 enzymes that are likely to collide with it. Compared to the previous situation of 1.vs.5, the 1.vs.3 situation here will have a lower rate.
Increasing the substrate concentration will increase reaction rate, two substrate molecules increase the chance of collision with the three enzymes. Even when the substrate molecules increase a lot, there will be only three enzymes working, saturated like before. Compared to 5 enzymes saturated, the rate of 3 enzymes working is lesser.

Therefore, the curve will be similar to the original reaction, except it will continuously be lower, even at the maximum rate, as explained above.
 
A

ashcull14

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qwertypoiu said:
Imagine there are 5 enzymes inside a solution.
Now if there was only one substrate molecule, after a while it will meet an enzyme and will form products.
Increasing the substrate concentration increases the number of molecules of the substrate per unit volume, thus increasing the number of likely collisions that happen in a given time initially, and thus increasing the rate of reaction. So, increasing substrate concentration increases the reaction rate. However, past a certain point the number of substrate molecules increases too much and the reaction rate no longer increases, this is because all five of the enzymes are saturated with a substrate to work on.

Now the inhibition part. A non-competitive inhibitor will just come in, and block let's say two of the enzymes. This means, when there is 1 substrate molecule, there is only 3 enzymes that are likely to collide with it. Compared to the previous situation of 1.vs.5, the 1.vs.3 situation here will have a lower rate.
Increasing the substrate concentration will increase reaction rate, two substrate molecules increase the chance of collision with the three enzymes. Even when the substrate molecules increase a lot, there will be only three enzymes working, saturated like before. Compared to 5 enzymes saturated, the rate of 3 enzymes working is lesser.

Therefore, the curve will be similar to the original reaction, except it will continuously be lower, even at the maximum rate, as explained above.
Click to expand...
tht was a superb example got it ;)
 
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